Let $V$ be a vector space over $\mathbb{C}$. Suppose $X\subset \mathbb{P} V$ is an algebraic variety, and consider its projective dual variety $X^\vee \subset \mathbb{P} V^*$. If the coordinate ring $\mathbb{C}[X]$ is arithmetically Cohen-Macaulay (aCM) is the same true for $\mathbb{C}[X^\vee]$?

An example where this is true is when $X$ is $n\times n$ matrices of rank $1$ and $X^\vee$ is $n\times n$ matrices of rank $n-1$.

I guess that it is true in general. Does someone have a reference (or a counter-example)?


  • $\begingroup$ I guess I should rephrase the question: What reasonable conditions on $X$ will guarantee that $X^\vee$ is aCM? $\endgroup$ Aug 25, 2014 at 16:19

1 Answer 1


Not true. Start from a variety $Y\subset \mathbb{P}V^*$ which is not aCM -- for instance a quartic rational curve in $\mathbb{P}^3$. In most cases, in particular in the example, its dual $X:=Y^\vee$ is a hypersurface in $\mathbb{P}V$, hence is aCM. But $X^\vee=Y$ is not.

  • 1
    $\begingroup$ And is there an example with $X$ smooth? $\endgroup$
    – Sasha
    Aug 21, 2014 at 8:33
  • $\begingroup$ Good question! It would have to be quite special, since the dual variety of "most" smooth varieties is a hypersurface. $\endgroup$
    – abx
    Aug 21, 2014 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.